Sobolev Embedding in $L^1$

Question

For $n\geq 1,$ is there a minimal $s$ (depending on $n$) such that $H^s(\mathbb{R}^n)$ is continuously embedded in $L^1(\mathbb{R}^n)?$ I researched several Sobolev embeddings books but I could not find such result or a counterexample. Thank you.

Answer 1

No. Sobolev embedding only increases the index p.

A typical example would be $(1+x^2)^{-1/2}$. It's $H^\infty$ but not integrable.